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A solution of urea (m ol. mass 56 g mol-1) boils at 100.18°C at the atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol-1 respectively, the above solution will freeze at [2005]
- a)0.654°C
- b)-0.654°C
- c)6.54°C
- d)-6.54°C
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A solution of urea (m ol. mass 56 g mol-1) boils at 100.18°C at th...
As ΔTf = Kf. m
ΔTb = Kb. m
ΔTb = Kb. m
Hence, we have 
= [ΔTb = 100.18 - 100 = 0.18°C]
As the Freezing Point of pure water is 0°C,
ΔTf = 0 –Tf
0.654 = 0 – Tf
∴ Tf = – 0.654 thus the freezing point of solution will be – 0.654°C.
ΔTf = 0 –Tf
0.654 = 0 – Tf
∴ Tf = – 0.654 thus the freezing point of solution will be – 0.654°C.
Most Upvoted Answer
A solution of urea (m ol. mass 56 g mol-1) boils at 100.18°C at th...
Given Data:
Molar mass of urea (M) = 56 g/mol
Boiling point of urea solution = 100.18°C
Kf for water = 1.86 K kg mol-1
Kb for water = 0.512 K kg mol-1
Calculation:
- The change in boiling point (ΔTb) can be calculated using the formula:
ΔTb = Kb * m
where m is the molality of the solution.
- Molality (m) can be calculated using the formula:
m = (moles of solute) / (mass of solvent in kg)
Since urea is the solute, moles of urea = given mass of urea / molar mass of urea
= 56g / 56 g/mol
= 1 mol
- Let's assume the mass of the solvent is 1 kg.
m = 1 mol / 1 kg
= 1 mol/kg
- Now, calculating the change in boiling point:
ΔTb = 0.512 * 1
= 0.512°C
- As the boiling point of the solution is 100.18°C, the new boiling point will be:
New boiling point = 100.18°C + 0.512°C
= 100.692°C
- The freezing point depression (ΔTf) can be calculated using the formula:
ΔTf = Kf * m
- Substituting the values:
ΔTf = 1.86 * 1
= 1.86°C
- The freezing point of the solution will be:
New freezing point = 0°C - 1.86°C
= -1.86°C
Therefore, the solution will freeze at -1.86°C, which is approximately -0.654°C. Hence, the correct answer is option B.
Molar mass of urea (M) = 56 g/mol
Boiling point of urea solution = 100.18°C
Kf for water = 1.86 K kg mol-1
Kb for water = 0.512 K kg mol-1
Calculation:
- The change in boiling point (ΔTb) can be calculated using the formula:
ΔTb = Kb * m
where m is the molality of the solution.
- Molality (m) can be calculated using the formula:
m = (moles of solute) / (mass of solvent in kg)
Since urea is the solute, moles of urea = given mass of urea / molar mass of urea
= 56g / 56 g/mol
= 1 mol
- Let's assume the mass of the solvent is 1 kg.
m = 1 mol / 1 kg
= 1 mol/kg
- Now, calculating the change in boiling point:
ΔTb = 0.512 * 1
= 0.512°C
- As the boiling point of the solution is 100.18°C, the new boiling point will be:
New boiling point = 100.18°C + 0.512°C
= 100.692°C
- The freezing point depression (ΔTf) can be calculated using the formula:
ΔTf = Kf * m
- Substituting the values:
ΔTf = 1.86 * 1
= 1.86°C
- The freezing point of the solution will be:
New freezing point = 0°C - 1.86°C
= -1.86°C
Therefore, the solution will freeze at -1.86°C, which is approximately -0.654°C. Hence, the correct answer is option B.
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A solution of urea (m ol. mass 56 g mol-1) boils at 100.18°C at the atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol-1 respectively, the above solution will freeze at [2005]a)0.654°Cb)-0.654°Cc)6.54°Cd)-6.54°CCorrect answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A solution of urea (m ol. mass 56 g mol-1) boils at 100.18°C at the atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol-1 respectively, the above solution will freeze at [2005]a)0.654°Cb)-0.654°Cc)6.54°Cd)-6.54°CCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solution of urea (m ol. mass 56 g mol-1) boils at 100.18°C at the atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol-1 respectively, the above solution will freeze at [2005]a)0.654°Cb)-0.654°Cc)6.54°Cd)-6.54°CCorrect answer is option 'B'. Can you explain this answer?.
A solution of urea (m ol. mass 56 g mol-1) boils at 100.18°C at the atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol-1 respectively, the above solution will freeze at [2005]a)0.654°Cb)-0.654°Cc)6.54°Cd)-6.54°CCorrect answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A solution of urea (m ol. mass 56 g mol-1) boils at 100.18°C at the atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol-1 respectively, the above solution will freeze at [2005]a)0.654°Cb)-0.654°Cc)6.54°Cd)-6.54°CCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solution of urea (m ol. mass 56 g mol-1) boils at 100.18°C at the atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol-1 respectively, the above solution will freeze at [2005]a)0.654°Cb)-0.654°Cc)6.54°Cd)-6.54°CCorrect answer is option 'B'. Can you explain this answer?.
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